What is 7+6i divided by 10+i?

Sep 8, 2017

$\frac{7 + 6 i}{10 + i} = \frac{76}{101} + \frac{53}{101} i$

Explanation:

We can make the denominator real by multiplying the denominator with its complex conjugate, thus:

$\frac{7 + 6 i}{10 + i} = \frac{7 + 6 i}{10 + i} \cdot \frac{10 - i}{10 - i}$

$\text{ } = \frac{\left(7 + 6 i\right) \left(10 - i\right)}{\left(10 + i\right) \left(10 - i\right)}$

$\text{ } = \frac{70 - 7 i + 60 i - 6 {i}^{2}}{100 - 10 i + 10 i - {i}^{2}}$

$\text{ } = \frac{70 + 53 i + 6}{100 + 1}$

$\text{ } = \frac{76 + 53 i}{101}$

$\text{ } = \frac{76}{101} + \frac{53}{101} i$

Sep 8, 2017

$\frac{76}{101} + \frac{53}{101} i$

Explanation:

$\frac{7 + 6 i}{10 + i}$
First we have to rationalize the denominator by multiplying the complex number in the denominator and the numerator by the denominator's conjugate.

$\frac{\left(7 + 6 i\right) \left(10 - i\right)}{\left(10 + i\right) \left(10 - i\right)} = \frac{7 \left(10\right) + 6 i \left(10\right) - 7 \left(i\right) - 6 i \left(i\right)}{{10}^{2} - {i}^{2}}$ (using the difference of squares rule in the denominator)
$= \frac{70 + 60 i - 7 i - 6 \left({i}^{2}\right)}{100 - {i}^{2}} = \frac{70 + 53 i - 6 \left(- 1\right)}{100 - \left(- 1\right)}$**(since ${i}^{2} = - 1$)

$\frac{70 + 53 i - 6 \left(- 1\right)}{100 - \left(- 1\right)} = \frac{70 + 6 + 53 i}{100 + 1} = \frac{76 + 53 i}{101}$

$= \frac{76}{101} + \frac{53}{101} i$

I hope that this helps.